Chemistry Academy

Target IIT - JEE

ORGANIC CHEMISTRY

ALKYL HALIDE
SHEET

Contents

Exercise I 2-6
Exercise II 7-9
Exercise III 10-13
Exercise IV 14-17
Exercise V 18-20
Answer key 20-25

Made by Neha Joshi

 EXERCISE-I
(One or more than one are correct)

Q.1 Which are possible products in following

moist Ag O


2


(A) (B) (C) (D)

Q.2 SN1 & SN2 is not favourable in

(A) H2C = CH–Cl (B) Ph–CH2–Cl (C) Ph–Cl (D) H2C=CH–CH2–Cl

Q.3 SN1 & SN2 product are same in (excluding stereoisomer)

(A) (B) (C) (D) Ph  CH  CH  CH 3

| |
CH 3 Cl

Q.4 Which of the following statements is / are true?

(A) CH3–CH2–CH2–I will react more readily than (CH3)2 CHI for SN2 reactions.
(B) CH3–CH2–CH2–Cl will react more readily than CH3–CH2–CH2–Br for SN2 reaction.
(C) CH3–CH2–CH2–CH2–Br will react more readily than (CH3)3C–CH2–Br for SN2 reactions
(D) CH3–O–C6H4– CH2Br will react more readily than NO2–C6H5–CH2Br for SN2 reaction

Q.5 A solution of (R)-2-iodobutane, [] = –15.9° in acetone is treated with radioactive iodied, until 1.0% of
iodobutane contain radioactive iodine, the specific rotation of recovered iodobutane is found to be
–15.58°, which statement is correct about above information.
(A) It has optical purity 96% (B) % of R in solution is 99%
(C) Racemic mixture is 2% (D) Final solution is dextrorotatory

Q.6 In which of the following case configuration about chiral C* is retained:

CH Br SOCl CH ONa
(A) Na
 3 (B) 2  3


(C) 
PCl
3  3
CH ONa
(D) H/
MeOH



Q.7 Which of the following will react with AcOAg

(A) R–NH2 (B) R–OTs
(C) R–N3 (D) R–Br

Alkyl Halides 2
Q.8 Consider the given reaction
CH 3
|
H  C  OTs NaCN
 CH 3CH 2 CH  CN
|
C 2 H 5 (S) |
CH 3
which of following statements are correct for above reaction.
(A) Product formation takes place due to the breaking of O–Ts
(B) The reaction is S 2
N
(C) The reaction is S N1
(D) Configuration of product is (R)

Q.9 Benzoyl chloride is less reactive than acetyl chloride for nucleophilic acyl substitution reaction because
O
||
(A)  C  group of benzoyl chloride is in conjugation with benzene ring. This makes benzoyl chloride
more stable than acetylchloride.
(B) CH3– group of acetyl chloride make C–Cl bond stronger due to the +I effect.
(C) C6H5– group of benzoyl chloride make C–Cl bond weaker due to +R effect.
(D) Carbonyl carbon in benzoyl chloride is less electron deficient than acetyl chloride

Q.10 SN2 reaction will be negligible in

Br

(A) (B) (C) (D)

Q.11 Rate of SN2 depends on

(A) Conc of Nucleophile (B) Conc of substrate
(C) Nature of leaving group (D) Nature of solvent

Q.12 Correct statement(s) for the product(s) of following reaction.

CH2 = CH – CH2 – Ph Cl  
/ 500C
2
(A) Four different products are formed
(B) Two optically active products are formed
(C) The optically active compound formed here can also be made by the reaction of HCl
(D) The reaction path is free radical substitution.

Q.13 In the given pair in which pair the first compound is more reactive than second to SN2 reaction.

Cl
(A) Cl CH2Cl (B)
Cl

Cl Cl

(C) (D)
Cl Cl

Alkyl Halides 3
Q.14 A gem dichloride is formed in the reaction :
(A) CH3CHO and PCl5 (B) CH3COCH3 and PCl5
(C) CH2 = CH2 and Cl2 (D) CH2 = CHCl and HCl

Q.15 Incorrect about alkyl halides is / are:

(A) Tertiary alkyl halides undergo SN2 substitutions
(B) Alkyl iodides on exposure to sunlight gradually darken
(C) Alkyl chlorides do not give beilstein test
(D) A nucleophilic substitution is most difficult in alkyl iodides


OH
80°C P+Q +R +S
Br

S
P.E. R
Q
Q.16
P
Br

Reaction co-ordinates
Correct statement(s) about reaction & related graph is (are):
(A) P & Q are nucleophilic substitution products by SN2 reaction mechanism
(B) only R & S are elimination products by E2 reaction mechanism
(C) P is nucleophilic substitution product by SN2 reaction mechanism
(D) Q,R,S are elimination products by E2 reaction mechanism

Q.17
which is / are correct statements about the product:

(A) is an endocyclic Saytzeff product

(B) is an exocyclic Saytzeff product

(C) is an exocyclic Hoffmann product

(D) is an endocyclic Hoffmann product

Alkyl Halides 4
Q.18 In which product formation takes place according to Hoffmann's rule
 
CH CH O K
(A) CH 3CH 2  CH  CH 3 t  B u O K (B) CH 3CH 2  CH  CH 3 
3 2

 
|  |
Br Br

 
OH OH
(C)   (D) CH 3CH 2CH  CH 3  
 
|

S(CH 3 ) 2

Q.19 Which of following will give syn-elimination

(A) CH3CH2CH2–OCOCH3 (B) CH 3CH 2CH  CH 3
|
Cl
CH3
|
(C) CH 3CH 2CH 2  O  C  S  CH 3 (D) C H 3C H 2C H 2 N  O 
|| |
S CH3
Q.20 Which of following are correct for given reaction




(A) Major product of reaction is (B) Major product is

(C) The reaction is thermal elimination reaction (D) The reaction is E2 reaction

Q.21 Which alkyl halide undergo E2 elimination

(A) 1 (B)

(C) (D)

Q.22 Which of the following reactions favors products at equilibrium.

(Assume that all reactants and products are soluble.)

(A) CH3Cl + I–  CH3I + Cl– (B) CH3Cl + F–   CH3F + Cl–

(C) CH3Cl + N3–  CH3N3 + Cl– (D) CH3Cl + –OCH3  CH3OCH3 + Cl+
(Hints: the pKa of HN3 is 4.72.)

Alkyl Halides 5
Q.23 The products of reaction of alcoholic silver nitrite with ethyl bromide are :
(A) Ethane (B) Ethene (C) Nitroethane (D) Ethyl nitrite

Q.24

The product(s) of this reaction can be

(A) (B)

(C) (D) none

Q.25 Which of the following is/are incorrectly matched

H
H 3C C l
SN
(A) Walden inversion
D 2

H
SN 1
H 3C Cl
(B) Walden inversion
D

C2H5
SN1
(C)
H5C6 Cl Only retention of configuration
CH3

C2H 5
SN1
H5C 6 Cl
(D) only inversion of configuration
CH 3

Alkyl Halides 6
 MATRIX
EXERCISE–II
Q.1 Match List I REACTANTS with List II PRODUCTS
List I List II

(A) CH3–O–SO2CH3 + CHO
2 5 (1) CH3–CH2–PH2

(B) CH3–CH2–I + PH3 (2) CH3–O–C2H5

 
(C) HC  C Na + CH3–CH2–Br (3) CH3–O–CH3

(D) CH3–Cl + CH3– O (4) CHC–CH2–CH3

Q.2 Match List-I with List-II for given S reaction

N2
Z–CH2Br + CH3O  Z–CH2–OCH3 + Br 
List-I (Z) List-II (relative reactivity)

(A) H– (P) 0.1

(B) CH3– (Q) 3

(C) C2H5– (R) 1

(D) (S) 100

Q.3 Match the List I (reaction) with List II (reaction intermediate)

List I List II

(A) CF3–CHCl2 alc /

.KOH

  CF2 = CCl2 (1) Transition state only
CH 3
| 
(B) CH  C  OH H
 CH 3  C  CH 2 (2) Carbocation
3
| |
CH 3 CH 3
alc.KOH
(C) CH3–CH2–Br   CH2=CH2 (3) Carbanion


Br
|
(D) CH 3  C  CH 3 aq
. KOH
 /
 CH 3  C  CH 2 (4) Free radical
| |
CH 3 CH 3

Alkyl Halides 7
Q.4 Match the List I with List II

List I (REACTION) List II (SUBSTRATE)

(A) E1CB (1) 3° Amine oxide

(B) Saytzeff alkene as major product (2) Xanthate

Cl
|
(C) E2 (3) CH 3  CH 2  CH  CH 3

(D) Ei (4) C6 H 5  CH 2  CH  CH 3
|
F

Q.5 Match the following.

Column I (reaction) Column II (type of reaction)


CH3CH2 O
(A)  (P) E1



CH3CH2 O
(B)  (Q) E2
CH3CH2OD

Ag O
(C) 
2
 (R) E1cb
moist

(D) alc
. KOH
 (S) Ei

Alkyl Halides 8
Q.6 Column I Column II
(Primary alkyl bromide) (SN2 relative rate)

(A) CH3–CH2–Br (P) 10–5

(B) Me–CH2–CH2–Br (Q) 10–2

(C) (R) 0.8

(D) (S) 1

Q.7 Column I Column II

Alkyl-P-toluene sulfonate Ethanolysis relative rate (50°C)

(A) CH3–CH2–OTs (P) 1010

(B) H2C = CH – CH2 – OTs (Q) 105

(C) Ph – CH2 – OTs (R) 400

(D) (S) 35

(E) Ph3C–OTs. (T) 1

Q.8 Substrate E2 elimination SN2-substitution

(A) CH3–CH2–Br (P) 1 (W)  0
(B) (CH3)2CH–Br (Q) 80 (X) 20
(C) (CH3)3CBr (R) 100 (Y) 90

Alkyl Halides 9
 COMPREHENSIONS
EXERCISE-III
PASSAGE – 1
A leaving group in an alkylating agent can be displaced not only by a nucleophile added to the reaction
mixture but also by one in the alkylating agent itself. This holds for alkylating agents that contain a
nucleophilic electron pair at a suitable distance from the leaving group. This neighboring group displaces
the leaving group stereoselectively through backside attack and this attack corresponds to the of an SN2
reaction. The nucleophilic electron pairs of the neighbouring group can be non bonding, or they can be in
a  bond. The higher rate of reversal substitution reactions are explained using this neighboring group
participation.

Q.1 (S)-2-chloro-3-ethyl-3-pentanol undergoes SN2 hydrolysis with aq. KOH to give

(A)(S)-3-ethylpentane -2,3-diol (B) (R)-3-ethylpentane -2,3-diol
(C)(R) -3-methylpentane -2,3-diol (D) (S) -3-methylpentane -2,3-diol

Q.2 The major product obtained in the following reaction is

aq.KOH
H 5C 2  S  CH 2  CH  Cl  
|
CH3

H5C2  S  CH 2  CH  OH H5C2  S  CH  CH 2OH

| |
(A) CH3
(B) CH3

(C) H 5C 2  S  C  CH 3  2 OH (D) H5C2  S  CH  CH  CH3

Q.3 The major product obtained in the given reaction is

Et OH 
N . ?
H 2O
Et Cl

.
Et
N .
(A) (B) Et O H
Et OH
N

Et

Et . .
(C) N (D)
Et OH

Alkyl Halides 10
PASSAGE – 2

R- L + Nu-  R – Nu + L-
The above nucleophilic substitution may be SN1 (or) SN1 reaction acemisation takes place. Where as
in SN2 reaction inversion occurs. When nucleophile is the isotope of leaving group racemisation occurs
n SN2 reaction Neighbouring group participation Br also possible in SN2 reaction

Q 4. In the following SN2 displacement reaction, 100% retention of configuration is observed. This is evidence
of
(A) Symmetrical intermediate (B) cationic intermediate
(C) Back side nucleophilic displacement (D) Neighbouring group participation.

Q. 5 Which of the following statements is correct about SN1 displacement reaction?

(A) Always racemic mixture is the product
(B) Along with racemic mixture , excess of inversion product is also formed
(C) Along with racemic mixture excess f retention product is also formed
(D) The rate of the reaction is independent of concentration of nucleophile.

Alkyl Halides 11
PASSAGE- 3
An organic compound A has molecular formula C5H9-Br. A decolourises brown colour of bromine
water but does not rotate plane polarized light. A on treatment with HBr/(PhCO)2O2 froms C5H10Br2
which on further treatment with Na/R2O gives a cyclic compound which on dibromination gives seven
isomeric products.

Q6 The starting compound A is most likely to be

Br
Br
(A) CH2 (B) H C
3
H3C

Br H 3C CH 2
(C) CH2
(D)
Br
Q 7. Compound A on treatment with HBr will produce
(A) An achiral dibromide (B) A racemic mixture
(C) A single pure enantiomer (D) A meso dibromide

Q8 The true statement regarding the monochloro derivative product formed in the reaction below is

H2/Ni Cl2
A Monochlorination
UV

(A)Three position isomers are formed (B) Total five str isomers are produced
(C)Two pair of enantiomers are formed (D) A pair of diastereomers are formed

Alkyl Halides 12
PASSAGE- 4

In the study of chlorination of propane, four products (A,B,C&D) (structural isomerism) of the formula
C3 H 6Cl2 were isolated. Each was further chlorinated to provide trichloro products  C3 H 5Cl3  .
It was found that A provide one trichloro product, B gave two and C&D each gave three. It is found that
D is optically active.

Q9 Formula of the compound A is

(A) (B)

(C) (D)

Q 10 Correct formula of the product of chlorination of B is —

(A) (B)

(C) Both A and B (D)

Q 11 Correct formula of the compound D is

(A) CH 3CCl2CH 3 (B) ClCH 2CH 2CH 2Cl
(C) CH 3CH 2CHCl2 (D) ClCH 2CHClCH 2

Alkyl Halides 13
 SUBJECTIVE
EXERCISE-IV

Q.1 Which nucleophile, :N(C2H5)3 or : P(C2H5)3, reacts most rapidly with methyl iodide in ethanol solvent?
Explain, and give the product formed in each case.

Q.2 Arrange following compounds according to their reactivity with alc silver nitrate.
t-Butyl chloride, sec butyl chloride and CCl4.

Q.3 Write major product of the the following reactions:

EtOH
(i) ClCH2CH2CH2Br + KCN  A
SF
(ii) PhCHO 
4 B
AgF
(iii) BrCH2CH = CHCO2Me 
 C
DMF
(iv) EtOH + HI  D
(v) EtOH + HCN  E

Q.4 Heating many alkyl chlorides or bromides in water effects their conversion into alcohol through a SN1
reaction. Order each of the following sets compounds with respect to solvolytic reactivity.
Br
Br
Br Br
(a)
(I) (II) (III)

Br
(b) Br
Br
(I) (II) (III)

Cl Cl Cl

(c)

Q.5 RCl is treated with Li in ether to form R – Li, R – Li reacts with water to form isopentane. R – Cl also
reacts with sodium to form 2, 7–dimethyloctane. What is the structure of R – Cl.

Q.6 Iodoform gives precipitate with AgNO3 on heating while CHCl3 does not.

Q.7 Explain why?

(a) Hydrogen atom of chloroform is definately acidic, but that of methane is not.
(b) A small amount of alcohol is added to chloroform bottles.
(c) RCl is hydrolysed to ROH slowly but reaction fastens on addition of KI.
(d) KCN reacts with R – I to give alkyl cyanide, while AgCN results in isocyanide as major product.

Q.8 The mechanism of decomposition of Me3 S+ OH– is SN2 whereas of Me3S+ I– is SN1.

Alkyl Halides 14
Q.9 Iodine reacts with alcohols to give alkyl iodine only in presence of phosphorous.

Q.10 Alkaline hydrolysis of benzyl chloride in 50% aqueous acetone proceeds by both SN2 and SN1 mechanism,
when water is used as solvent, mechanism was now mainly SN1.

Q.11 On electrolysis of aqueous ethanolic solution of sodium chloride gives sweet smelling liquid (X). (X)
gives isocyanide test and condenses with acetone to from hypnotic (Y). What are (X) and (Y)?

Q.12 Predict the product(s) and write the mechanism of each of the following reactions
HI (1mole) excess HI
(i)  (ii) 
O O

Q.13 A chloroderivative ‘X’ on treatment with zinc and hydrochloric acid gave a hydrocarbon with five carbon
atoms in the molecule. When X is dissolved in ether and treated with sodium, 2, 2, 5, 5-tetramethyl
hexane is obtained. What is compound X.

Q. 14 A hydrocarbon C8H10 (A) on ozonolysis gives compound C4H6O2 (B) only. The compound (B) can
also be obtained from the alkyl bromide C3H5Br (C) upon treatment with magnesium in dry ether
followed by CO2 and acidification. Identify (A), (B) and (C) and also give equations for the reactions.

Q.15 In study of chlorination of propane four products (A,B,C,D) of molecular formula C3H6Cl2 were obtained.
On further chlorination of the above products A gave one trichloro product, B gave two whereas C and
D gave three each. When optically active C was chlorinated one of trichloro propanes was optically
active and remaining two were optically inactive. Identify the structures of A,BC and D, and explain
formation of products.

Q.16 Identify A,B,C,D,E and F in the following series of reaction.

Br
2  A aq
. KOH
 Na
 B  C
h

alc.KOH

C
D NBS
 E  F

Q.17 What are the products of the following reactions?

CH 3 CH 3
| |
(a) CH 3  C  Cl  OCH 3  (b) CH 3  C  O ¯ + CH3 – X 
| |
CH 3 CH 3

Q.18 Complete the following by providing the structure of (A), (B), (C) and (D):
Alc.KOH HBr NH 3
(i) CH3CH2CH2OH PBr
3  (A)   (B)  (C)  (D)

H / H 2O SOCl H
(ii) CH3CH2CH2I Alc
. KOH
 (A)    (B) 2  (C)  (D)
LiAlH4

Alc.KOH HBr
(iii)  (A)   (B) 
CH3CH2CH = CH2 NBS
  (C)
Light
HBr
(iv) CH3CH2MgBr CH CHO / H 2O
3    (A)  (B) Alc
. KOH
 (C)

Alkyl Halides 15
Q.19 CH 3–CH2I reacts more rapidly with strong base in comparison to CD3CH2I

Q.20 (a)

(b)

What are (A) & (B) and explain their stereochemistry.

Q.21 Propose a mechanism for the following reactions-

CH3 OH
CH3
C—Br H
2O
 
CH3
CH3

Q.22 Each of the following alcohols has been subjected to acid catalyzed dehydration and yields a mixture of
two isomeric alkenes. Identify the two alkenes in each case, and predict which one is the major product
on the basis of the Zaitsev rule.
OH

(a) (CH 3 ) 2 CCH (CH 3 ) 2 (b) (c)

|
OH
H
Q.23 Give the major product (with proper explanation) when following halogen compounds are treated with
sodium ethoxide.
CH2Br
(a) CH 3  CH  CHCH 3 (b) (c) CH3
| | CH3
Br CH 3
Cl
Q.24 2-chloro-3-methylbutane on treatment with alcoholic potash gives 2-methylbutene-2 as major product.
Explain.

Q.25 CHC–CH2–CH=CH2, adds up HBr to give CHC–CH2–CHBr–CH3 while CHC–CH=CH2 adds

up HBr to give CH2=C . Br . CH=CH2.

Alkyl Halides 16
Q.26 A primary alkyl bromide (A), C4H9Br, reacted with alcoholic KOH to give compound (B). Compound
(B) reacted with HBr to give an isomer of (A), (C). When (A) was reacted with sodium metal it gave
compound (D), C8H18, which was different from the compound produced when n-butyl bromide was
reacted with sodium. Draw the structure of (A) and write equations for all the reactions.

Q.27 Carry out following conversions.

(a) CH3–CH2–CH2–Cl — CH 3  CH  CH 3 in single step
|
Cl

(b) — in three steps only

(c) — in three steps only

Q.28 Convert

(i) CH2 — CH3  CH = CH2

(ii) CH3CH2CH=CH2 CH3CH2CH2CH2NH2

(iii) CH3CH2CH=CH2  CH2=CH—CH=CH2

(iv) OH–CH2–CH2CH = CH2 

O
Q.29 Carry out following conversions.
Cl
|
(a) CH 3  CH  CH 3 — CH3–CH2–CH2–Cl without using peroxide

O
||
(b) Ph  CH 2  C  Cl — Ph – CH2 – Cl

O
||
(c) Ph – CH2 – Cl — Ph  CH 2  C  Cl

Q.30 Treatment of 2-bromobutane with hot alcoholic KOH gives a mixture of three isomeric butenes (A), (B)
and (C). Ozonolysis of the minor product (A), gives formaldehyde and another aldehyde in equimolar
amounts. What are the structural formulae of (A), (B) and (C)?

Alkyl Halides 17
 INTEGER
EXERCISE–V
Q.1 Choose the alkyl halide(s) from the following list of C6H13Br isomers that meet each criterion below.
(1) 1 -bromohexane (2) 3-bromo-3-methylpentane

(3) 1 -bromo-2,2-dimethylbutane (4) 3-bromo-2-methylpentane

(5) 2-bromo-3-methylpentane

(A) the compounds(s) that can exist as enantiomers

(B) the compounds(s) that can exist as diastereomers

(C) the compounds that gives the fastest S^ reaction with sodium methoxide

(D) the compound that is least reactive to sodium methoxide in methanol

(E) the compounds(s) that give only one alkene in the E2 reaction

(F) the compounds(s) that give an E2 but no S^ reaction with sodium methoxide in methanol

(G) the compound(s) that undergo an SN1 reaction to give rearranged products

(H) the compound that gives the fastest SN1 reaction

Q.2 Of the following statements how many are true for SN1 reaction.

(a) Tertiary alkyl halides react faster than secondary.

(b) The absolute configuration of the product is opposite to that of the reactant when an optical
active substrate is used.

(c) The reaction shows first order kinetics.

(d) The rate of reaction depends markedly on the nucleophilicity of the nucleophile.

(e) The mechanism is two step.

(f) Carbocations are intermediate.

(g) Rate  [Alkyl halides]

(h) The rate of the reaction depends on the nature of the leaving group.

Alkyl Halides 18
Q.3 Of the following statements.how many are true for SN2 reaction.

(a) Tertiary alkyl halides reacts faster than secondary.

(b) The absolute configuration of product is opposite to that of the reactant when an optically active
substrate is used.

(c) The reaction shows first order kinetics.

(d) The rate of the reaction depends markedly on the nucleophilicity of the attacking reagent.

(e) The mechanism is one step.

(f) Carbocations are intermediate.

(g) Rate  [Alkyl halides]

(h) The rate of the reaction depends on the nature of the leaving group.

Q. 4 How many of the following give exclusively isobutene on dehydrohalogenation ?

i) n - butyl chloride ii) tertiary butyl bromide
iii) isobutyl chloride iv) Secondary butyl chloride

Q5 Find the total number of possible monobrominate product obtained during the reaction

Q6 The total number of alkenes possible by dehydrobromination of 3-bromo-3 cyclopentylhexane using

alcoholic KOH is
Q. 7 The total number(s) of stable conformers with non-zero dipole moment for the following compound
is(are)

Alkyl Halides 19
Q.8

x = Total number of substitution & elimination product(s). Find the value of X?

Q.9 How many of the following reactions will prefer following a path way which give 1st order kinetics.

Q.10. x = number of compound undergo nucleophilic substitution reaction

Alkyl Halides 20
 ANSWER KEY
One of more than one correct
EXERCISE-I
Q.1 A,B Q.2 A,C Q.3 B,C Q.4 A,C
Q.5 B,C Q.6 A,C Q.7 B,C,D Q.8 B,D
Q.9 D Q.10 A,B,C Q.11 A,B,C,D Q.12 A,B,D
Q.13 B,D Q.14 A,B,D Q.15 A,C,D Q.16 C,D
Q.17 A Q.18 A,C,D Q.19 A,C,D Q.20 A,C
Q.21 A,B Q.22 B,C,D Q.23 C,D Q.24 A,B,C
Q.25 B,C,D

Matrix
EXERCISE–II

Q.1 (A) 2 ; (B) 1 ; (C) 4 ; (D) 3 Q.2 (A) S ; (B) Q ; (C) R ; (D) P
Q.3 (A) 3 ; (B) 2 ; (C) 1 ; (D) 2 Q.4 (A) 4 ; (B) 3, 4 ; (C) 3 ; (D) 1, 2
Q.5 (A) Q ; (B) R ; (C) P ; (D) Q Q.6 (A)S ; (B)R ; (C)Q ; (D)P
Q.7 (A)T ; (B)S ; (C)R ; (D)Q ; EP Q.8 (A)P,Y ; (B)Q, X ; (C) R,W

Comprehension
EXERCISE–III

Q.1 A Q.2 B Q.3 B Q.4 D Q.5 B

Q.6 C Q.7 B Q.8 B Q.9 B Q.10 C
Q.11 D

Subjective
EXERCISE-III
Q.1 P(Et)3
Q.2 tert-Butyl chloride, sec butyl chloride and CCl4 with alc. silver nitrate

Me3C – Cl > CHCl > CCl4

Q.3 (i) ClCH2CH2CH2CN (ii) PhCHF2

(iii) FCH2CH = CHCO2Me (iv) EtI
(v) no reaction
Q.4 (a) III > II > I (b) III > I > II (c) II > I > III

Q.6 C–I bond being less stable than C–Cl bond and thus on heating heterolytic cleavage of C –I form I–
which gives yellow precipitate with AgNO3

Q.7
(a) Due to three electronegative chlorine atoms present on carbon, the latter acquires a partial positive
charge due to –I effect of chlorine. As a result, it tends to attract electrons of the C–H bond towards
itself. Hence the removal of hydrogen atom as proton becomes easier.
(b) Alcohol acts as inhibitor for oxidation of chloroform. Also it reacts with COCl2 to give harmless diethyl
carbonate.

Alkyl Halides 21
(c) KI reacts with RCl to form RI. This RI molecule now hydrolysed easily to give ROH because alkyl
iodide are more reactive than alkyl chloride.
RCl + KI  RI HOH  ROH (faster)
(d) KCN is an ionic compound [K+(:C N:)–] in which both C and N carry a lone pair electron. Carbon carrying
lone pair of electrons is more reactive and thus alkyl attacks carbon to give alkyl cyanideAgCN being covalent
has Ag– CN : structure with lone pair on N thus R attacks on N atom and R–N   C is formed.

Q.8 More the nucleophilic activity of the attacking reagent, more will be the SN2 path. Since in SN1 mechanism
the raegent doesnot enter into the rate determining step of ionisation. However it can also be expected
that lener nucleophilic activtiy changes the mechanism from SN2 to SN1
Q.9 Phosphorus reacts with I2 to give PI3 which replaces OH group of alcohol to produce R – I
2P + 3I2  2 PI3
3R OH + PI3  3RI + H3PO3
Q.10 The dielectric constant of water is greater than that of aqueous acetone, and so ionisation of benzyl
chloride is facilitated.

HI(1 mole) I
Q.12
 
O O I
OH
H

HI(XS) HI
 –H2O
O I I I
OH
CH3

Q.13 x is CH3—C—CH2—Cl

CH3

Mg (i) CO2 ozonolysis

Q. 14 Br ether
MgBr (ii) H3O
C—OH CC
O
Cyclopropyl (C) Cyclopropane (B) Dicyclopropyl (A)
bromide carboxylic acid acetylene

Cl
Cl
Q.15 (A) CH3—C—CH3 (B) CH2—CH2—CH 2 (C) CH3—CH—CH 2 (D) CH 3—CH 2—CH
Cl
Cl Cl Cl Cl Cl

Alkyl Halides 22
  
Br OH O Na

Br2 aq. KOH Na

h
(A) (B) (C)
alc. KOH

NBS (C)
Q.16 Br
(D) (E) (F)
O

Q.17 (a) (b) OCH3

Q.18 (i)A, CH3CH2CH2Br ; B, CH3CH = CH2 ; C, CH3CHBrCH3; D, CH3CHNH2CH3

(ii) A, CH3CH = CH2; B, CH3CHOHCH3; C, CH3CHClCH3; D, CH3CH2CH2
(iii) A, CH3CHBrCH = CH2; B, CH2=CH–CH=CH2; C, CH3CHBrCH=CH2& CH3CH = CH–CH2Br
(iv) A, CH3CH2CHOHCH3; B, CH3CH2CHBrCH3; C, CH3CH = CH – CH3

Q.19 The elimination of HI (or DI) in presence of strong base shows E2 elimination. The rate determining step
involves breaking up of C – H (or C–D) bond. The C–D bond being stronger than C–H and thus
elimination is faster in case of CH3 – CH2I.

Q.20 (a) A  B C D

E F G H I

Me
(b) (A) , (B) Pr OEt
H
Q.22 Stability of alkene by -hydrogen
CH2OCH2CH3
Q.23 (a) (b) (c)

Q.24 H 3C  CH  CH  CH 3 KOH
 ( alc
)  CH 3  C  CH  CH 3  CH 3  CH  CH  CH 2
| | | |
CH 3 Cl CH 3 CH 3
major minor
Elinination occurs according to saytzeff rule. The major product is one which involves elimination of H
from less hydrogenated carbon.

Alkyl Halides 23
 CH3 CH3
alc KOH
CH3—C—CH2—Br CH3—C = CH2
Q.26 A HBr
H
CH3
CH3—C – CH3
Br
Na / DE
CH3 CH3
H3C— C — C —CH3
CH3 CH3

Cl
Q.27 (a) CH3—CH—CH3 alc KOH B2H6
CH3—CH = CH2 H3C—CH2—CH2—Cl
Cl2
Cl
AlCl3 |
(b) CH3—CH2—CH2—Cl CH3—CH—CH3

Br
CH 2— CH2 OH Br
alc KOH O3/LAH red P/Br3
(c)

Cl
CN + CH2—OH CH2—Cl
(d) KCN H3O /LAH SOCl2

O
NaN3/HO
(e) Ph—CH2—C—Cl Ph—CH2—NH2 NOCl Ph—CH2—Cl

H3O+
(f) Ph—CH2—Cl KCN Ph—CH2—CN Ph—CH2—COOH
SOCl2
O
Ph—CH2—C—Cl

CH2—CH3 NBS alc KOH

Q.28 (i) CH—CH3 CH = CH2

Br
(ii) (i) B2H6/NH2—Cl
CH3—CH2—CH = CH2 CH3—CH2—CH2—CH2—NH2

(iii) CH3—CH2—CH = CH2 NBS CH3—CH—CH = CH2 alc KOH 1 = C—C = C

Br

Alkyl Halides 24
 B2H6/Cl2
(iv) HO—CH2—CH2—CH = CH2 HO—CH2—CH—CH—CH2

Na Cl

Q.30 (A) (B) (C)

INTEGER
EXERCISE–IV
ANSWERS
Q.1 (a) Compounds (4) and (5) can exist as enantiomers.
(b) Compound (5) can exist as diastereomers because it has two asymmetric carbons.
(c) Compound (1) gives the fastest SN2 reaction with sodium methoxide because it is the only primary
alkyl halide
(d) Compound (3) is least reactive to sodium methoxide in methanol because it cannot undergo a
-elimination it has no -hydrogens - and the three -branches make it virtually unreactive in the SN2
reaction, much like neopentyl bromide.
(e) Compound (1) can give only one alkene in the E2 reaction.
(f) Compound (2) will give an E2 but no SN2 reaction with sodium methoxide in methanol.
(g) Compound (4) and (5) undergo the SN1 reaction with rearrangement If compound (3) is forced to
react, it will also undergo the SN1 reaction with rearrangement, but under ordinary conditions it will not
react.
(h) Compound (2) will give the fastest SN1 reaction because it is the only tertiary alkyl halide.
Q.2 6 a, c, e, f, g, h
Q.3 5 b, d, e, g, h
Q. 4 2 Q. 5 4 Q. 6 5 Q. 7 3 Q. 8 4
Q. 9 3 Q. 10 6

Alkyl Halides 25